Problem: Complete the equation. $\dfrac38+ \dfrac38 + \dfrac38~=~$
Answer: Let's figure out what $\dfrac{3}{8} + \dfrac{3}{8} + \dfrac{3}{8}$ equals. $\dfrac{0}{8}$ $\dfrac{3}{8}$ $\dfrac{6}{8}$ $\dfrac{9}{8}$ $\llap{{+}}\!\frac{3}{8}$ $\llap{{+}}\!\frac{3}{8}$ $\llap{{+}}\!\frac{3}{8}$ $\dfrac{3}{8} + \dfrac{3}{8} + \dfrac{3}{8} = \dfrac{9}{8}$ Now, let's figure out how many times we add $\dfrac{1}{8}$ to make $\dfrac{9}{8}$. $\dfrac{0}{8}$ $\dfrac{1}{8}$ $\dfrac{2}{8}$ $\dfrac{3}{8}$ $\dfrac{4}{8}$ $\dfrac{5}{8}$ $\dfrac{6}{8}$ $\dfrac{7}{8}$ $\dfrac{8}{8}$ $\dfrac{9}{8}$ $\llap{{+}}\!\frac{1}{8}$ $\llap{{+}}\!\frac{1}{8}$ $\llap{{+}}\!\frac{1}{8}$ $\llap{{+}}\!\frac{1}{8}$ $\llap{{+}}\!\frac{1}{8}$ $\llap{{+}}\!\frac{1}{8}$ $\llap{{+}}\!\frac{1}{8}$ $\llap{{+}}\!\frac{1}{8}$ $\llap{{+}}\!\frac{1}{8}$ $=\overbrace{{\dfrac1{8}} +{\dfrac1{8}} +{\dfrac1{8}} +{\dfrac1{8}} +{\dfrac1{8}} + {\dfrac1{8}} +{\dfrac1{8}} +{\dfrac1{8}} + {\dfrac1{8}}}^{{9}\text{ eighths}} $ $=\dfrac{{9}\times{1}}{{8}}$ $\dfrac{3}{8}+ \dfrac{3}{8} + \dfrac{3}{8} = 9 \times \dfrac18$